What would be the wavelength and name of series respectively for the emission transition for H-atom if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm?

Options

A. 434 nm, Balmer
B. 434 pm, Paschen
C. 545 pm, Pfund
D. 600 nm, Lyman

Answer

A

Explanation

Radius of `n^(th)` orbit
`=(0.529xxn^2)/(Z)Å =(52.9xxn^2)/(Z)` pm
Radius `(r_2)=211.6 " pm " = (52.9n_1^2)/(z)`
`therefore r_2/r_1=(1322.5)/(211.6)=(n_2^2)/(n_1^2)`
`rArr 6.25=(n_2^2)/(n_1^2)=(n_2/n_1)^2 rArr n_2/n_1= sqrt(6.25)=2.5`
`therefore n_1=2, n_2=5` thus, the transition is from `5^(th)` orbit to `2^(nd)` orbit. it belong to Balmer series.
`barv=R(1)/(n_1^2)-1/(n_2^2) rArr barv= 1.097xx10^7xx21/100`
`lamda = 1/(barv)=(100)/(1.097xx21xx10^7)m`
`=4.34xx10^(-7)m=434xx10^(-9)` m
`lamda=434nm`
Thus, it lies in the visible region.